Respuesta :
Answer
given,
Mass of the runner, M = 70 Kg
speed of the runner on the second base = 4.35 m/s
speed at the base = 0 m/s
Acceleration due to gravity,g = 9.8 m/s²
a) magnitude of mechanical energy lost
Mechanical energy lost is equal top gain in kinetic energy
[tex]ME_{Lost}=\dfrac{1}{2}mv^2[/tex]
[tex]ME_{Lost}=\dfrac{1}{2}\times 70\times 4.35^2[/tex]
[tex]ME_{Lost}=662.29\ J[/tex]
b) Work done = Force x displacement
W = F. x
F = μ mg
W = μ mg . x
Work done is equal to 662.29 J
[tex]x=\dfrac{W}{\mu m g}[/tex]
using the coefficient of the friction,μ = 0.7
[tex]x=\dfrac{662.29}{ 0.7\times 70\times 9.8}[/tex]
x = 1.38 m
Hence, the runner will slide to 1.38 m.
(A) The magnitude of the mechanical energy lost due to friction acting on the runner is 662.28 J.
(B) The distance covered by runner during the sliding by friction is 0.22 m.
Given data:
The mass of base runner is, [tex]m=70 \;\rm kg[/tex].
The initial speed of runner is, [tex]u =4.35 \;\rm m/s[/tex].
The final speed is, [tex]v=0 \;\rm m/s[/tex].
The acceleration due to gravity is, [tex]g =9.8 \;\rm m/s^{2}[/tex].
(A)
The mechanical energy lost by the runner due to friction is equal to the magnitude of kinetic energy of runner. Then,
Mechanical energy lost = Kinetic energy.
[tex]ME = KE\\ME=\dfrac{1}{2}mu^{2}\\ME=\dfrac{1}{2} \times 70 \times 4.35^{2}\\\\ME = 662.28 \;\rm J[/tex]
Thus, we conclude that magnitude of the mechanical energy lost due to friction is 662.28 J.
(B)
Apply the third kinematic equation of motion to obtain the distance covered by runner as,
[tex]v^{2}=u^{2}+2(-g)h[/tex]
Here, h is the distance covered by runner.
Solving as,
[tex]0^{2}=4.35^{2}+2(-9.8)\times h\\h=\dfrac{4.35}{2 \times 9.8}\\h=0.22 \;\rm m[/tex]
Thus, the distance slide by the runner is 0.22 m.
Learn more about kinematic equations of motion here:
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